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2 years ago

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An accelerometer is a device that uses the extension of a spring to measure acceleration in terms of Earth's gravitational accel

eration (g). What is the approximate acceleration if this accelerometer spring is extended just 0.43 centimeters?. . a) 0.24 g. b) 0.80 g. c) 2.00 g. d) 2.40 g

2 answers:

Nimfa-mama [501]2 years ago

8 0

0.80g (B) is correct for plato!

ELEN [110]2 years ago

3 0

In your question explains that an accelerometer is a device that uses the extension spring to measure acceleration in terms of Earth's Gravitational accelerations. So if this device is extended just 0.43 centimeters the possible accelerations is letter B. 0.80 g

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(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

avanturin [10]

6m at 0.9 radian means 6m at $51.57^0$

Since positive angles are counter clockwise

6m at $51.57^0$ can be written as 6*cos $51.57^0$ i+6*sin $51.57^0$ j = 3.73i+4.70j

5m at $-75^0$ can be written as 5*cos $(-75)^0$ i+5*sin $(-75)^0$ j = 1.294i-4.83j

4m at 1.2 radian means 4m at $68.75^0$

4m at $68.75^0$ can be written as 4*cos $68.75^0$ i+4*sin $68.75^0$ j = 1.45i+3.73j

6m at $-210^0$ can be written as 6*cos $(-210)^0$ i+6*sin $(-210)^0$ j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = $\sqrt{1.274^2+0.6^2} = 1.98 m$

c) Angle, $tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0$

7 0

2 years ago

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26× $10^{13}$

==> $v^{2}$ = 1.1 × $10^{19}$ ÷ 6.26× $10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5× $10^{8}$ ÷ 0.419188323 × $10^{3}$

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0

2 years ago

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A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

valentina_108 [34]

F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer

7 0

2 years ago

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A swimming pool heater has to be able to raise the temperatureof the 40,000 gallons of water in the pool by 10 degress Celsius.H

andrew11 [14]

Answer:b)1770 kWh

Explanation:

Given

volume of water $V=40,000 gallons$

Temperature rise $\Delta T=10^{\circ}C$

$c=4186 J/kg-^{\circ} C$

also 1 kg mass is approximately is 1 gallon

therefore 40,000 gallon is equivalent to 3.8\times 40000 kg

heat Required to raise temperature is

$Q=mc\Delta T$

$Q=3.8\times 40000\times 4186\times 10$

$Q=63,627.2\times 10^5 J$

$Q=63,672.2\times 10^2 KJ$

$Q=1767.42 kWh\approx 1770 kWh$

3 0

2 years ago

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

2 years ago